How many roots of the equation `3x^4+6x^3+x^2+6x+3=0` are real ?

Explanation

To determine how many roots of the equation \(3x^4 + 6x^3 + x^2 + 6x + 3 = 0\) are real, we can follow these steps:

### Step 1: Analyze the Equation
We start with the polynomial equation:
\[ 3x^4 + 6x^3 + x^2 + 6x + 3 = 0 \]

### Step 2: Divide by \(x^2\)
To simplify the equation, we divide every term by \(x^2\) (assuming \(x \neq 0\)):
\[ 3x^2 + 6x + 1 + \frac{6}{x} + \frac{3}{x^2} = 0 \]
This can be rewritten as:
\[ 3x^2 + 6x + 1 + 6 \cdot \frac{1}{x} + 3 \cdot \frac{1}{x^2} = 0 \]

### Step 3: Substitute \(t = x + \frac{1}{x}\)
Let \(t = x + \frac{1}{x}\). Then, we can express \(x^2 + \frac{1}{x^2}\) in terms of \(t\):
\[ x^2 + \frac{1}{x^2} = t^2 - 2 \]
Substituting this into the equation gives:
\[ 3(t^2 - 2) + 6t + 1 = 0 \]
This simplifies to:
\[ 3t^2 + 6t - 5 = 0 \]

### Step 4: Solve the Quadratic Equation
Now, we can solve the quadratic equation \(3t^2 + 6t - 5 = 0\) using the quadratic formula:
\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \(a = 3\), \(b = 6\), and \(c = -5\):
\[ t = \frac{-6 \pm \sqrt{6^2 - 4 \cdot 3 \cdot (-5)}}{2 \cdot 3} \]
Calculating the discriminant:
\[ t = \frac{-6 \pm \sqrt{36 + 60}}{6} \]
\[ t = \frac{-6 \pm \sqrt{96}}{6} \]
\[ t = \frac{-6 \pm 4\sqrt{6}}{6} \]
\[ t = -1 \pm \frac{2\sqrt{6}}{3} \]

### Step 5: Determine the Range of \(t\)
The values of \(t\) are:
\[ t_1 = -1 + \frac{2\sqrt{6}}{3}, \quad t_2 = -1 - \frac{2\sqrt{6}}{3} \]
Next, we need to check if these values satisfy the condition for \(t = x + \frac{1}{x}\).

### Step 6: Analyze the Values of \(t\)
1. The expression \(x + \frac{1}{x} \geq 2\) for \(x > 0\) and \(x + \frac{1}{x} \leq -2\) for \(x < 0\).
2. We need to check if \(t_1\) and \(t_2\) fall within these ranges.

### Step 7: Count the Real Roots
- If \(t_1\) is greater than or equal to 2, it corresponds to two positive roots.
- If \(t_2\) is less than or equal to -2, it corresponds to two negative roots.

After evaluating \(t_1\) and \(t_2\), we find that:
- \(t_1\) is less than 2.
- \(t_2\) is less than -2.

Thus, we conclude that there are **two real roots** for the original equation.

### Final Answer
The equation \(3x^4 + 6x^3 + x^2 + 6x + 3 = 0\) has **2 real roots**.

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